The graph of the curve $x^{2} + y^{2} - 2 x y - 8 x - 8 y + 32 = 0$ falls wholly in the

first quadrant

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second quadrant

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third quadrant

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none of these

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Solution

The correct option is D first quadrant
$x^{2} + y^{2} - 2 x y - 8 x - 8 y + 32 = 0$
$\Rightarrow (x - y)^{2} = 8 (x + y - 4)$
is a parabola whose axis is $x - y = 0$ and the tangent at the vertex is $x + y - 4 = 0$ .
Also, when $y = 0$ , we have
$x^{2} - 8 x + 32 = 0$
which gives no real values of $x$ .
when $x = 0$ , we have $y^{2} - 8 y + 32 = 0$ which gives no real values of $y$ .
So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.

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The graph of the curve x2+y2−2xy−8x−8y+32=0 falls wholly in the

The graph of the curve $x^{2} + y^{2} - 2 x y - 8 x - 8 y + 32 = 0$ falls wholly in the